Page 221 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

208 Solutions to the Exercises

Exercise 4.3.2. This is a classical result.
(i) We start by choosing ψ ∈ C ∞ (0, 1), ψ ≥ 0,so that
0
Z 1 1
r n−1 ψ (r) dr = (7.18)
0 σ n−1

where σ n−1 = meas (∂B 1 (0)).We then let ψ ≡ 0 outside of (0, 1) and we define
for every > 0
µ ¶
1 |x|
ϕ (x)= ψ .
n

n o
n
(ii) Let Ω = x ∈ R : B (x) ⊂ Ω .Let x ∈ Ω , the function y → ϕ (x − y)

has then its support in Ω since supp ϕ ⊂ B (0). We therefore have

Z Z Z µ ¶
1 |y|
u (y) ϕ (x − y) dy = u (x − y) ϕ (y) dy = n u (x − y) ψ dy

R n R n |y|<
Z Z 1 Z
= u (x − z) ψ (|z|) dz = u (x − ry) ψ (r) r n−1 drdσ (y) .
|z|<1 0 |y|=1
(7.19)
We next use the mean value formula
Z Z
1 1
u (x)= udσ = u (x + ry) dσ (y) . (7.20)
σ n−1 r n−1 ∂B r (x) σ n−1 |y|=1
Returning to (7.19) and using (7.18) combined with (7.20), we deduce that

Z
u (x)= u (y) ϕ (x − y) dy .

R n
n
Since ϕ ∈ C ∞ (R ), we immediately get that u ∈ C ∞ (Ω ).Since is arbitrary,
0
we find that u ∈ C ∞ (Ω), as claimed.
α
Exercise 4.3.3. Let V (r)= |log r| and
u (x 1 ,x 2 )= x 1 x 2 V (|x|) .

A direct computation shows that

2 2
x x 2 x 1 x 2
1
0
= x 2 V (|x|)+ 0 = x 1 V (|x|)+ V (|x|)
u x 1 V (|x|) and u x 2
|x| |x|